Sealed CO2 power supply testing
Neil Baylis
brand new, never been powered up since I got it. Currently, I want to
test it, without a laser tube. (The tube will come later).
The data sheet with the PSU is kind of difficult to understand, but
seems to imply that it's a current regulator. It says it has a max
voltage of 30kV and max current of 30mA. There's a voltage input that
I believe sets the operating current for a particular tube, and a PWM
input to modulate the power.
Questions:
1. Is there any reason I shouldn't test this with a pure resistive
load? I was thinking of stringing a bunch of power resistors together
with a 100mA meter in series.
2. Looking on eBay at a typical 40 watt tube, it says operating
voltage 15 kV @ 18 mA. By my math, if I was to set up a dummy load to
mimic such a tube, it would need to dissipate 270 Watts and have a
resistance of 833k. I don't know if my PSU will do 40 watts, but I'm
pretty sure it should do at least 25. (I've lost my records from when
I ordered it).
How would I go about characterizing this power supply? The
instructions are adamant that it should not be run either shorted or
open.
Lostgallifreyan
7dfced...@x5g2000prf.googlegroups.com:
First thing is to read and think and sit and watch wait for better replies,
etc.. and do anything other than try it till you have no further reason NOT
to do so, and you might find a proper laser tube load before you try this PSU
some other way.
If you have 30 mA being pushed around by 30 kV you have such a high risk of
immediate painful death that playing Russian Roulette by dodging the
raindrops from a lawn sprinkler spurting concentrated H2SO4 can be considered
the soft option! There is no way to overstate this case so I might as well
wax lyrical about it. No matter what you know about HV, reconsider it deeply
before you try that thing.
Secondly, if you do try it then a resistive load makes sense but you have to
be sure that it won't change its character too much to allow a 'control' to
gauge the PSU with. If the resistor burns (most will with that supply across
them), then you might as well just destroy it with a short cicuit, against
existing advice. If you're really set on trying this, then I suggest using a
series chain of solid carbon composite resistors. While they vary a lot of
value under any conditions let alone these, they won't burn so fast you can't
react in time, and likely won't burn at all if you calculate right. Make the
network on a clean dry slate or ceramic slab. Trying to suspend it in air
might be electrically safe IF it holds under stress, but physically so
accident prone I wouldn't dare try it.
When you say 'do 40W' I take it you mean it will do enough to push 40W out of
a tube, plus whatever gets lost from that which gets pushed in? Anyway, Sam#s
LaserFAQ has this page:
http://www.repairfaq.org/sam/laserco2.htm
Prowl around that for some time...
STEVE ROBERTS
>
> How would I go about characterizing this power supply? The
> instructions are adamant that it should not be run either shorted or
> open.
The answers are,
1. You DONT. running open or shorted will KILL the unit. Running
without a plasma load of some sort will kill the unit, as will any
ARCING.
The Chinese switchers simply do not have the protection systems a
western power supply does.
2. You could test it on a neon tube about 3 foot long with a 50-80K 5
watt series wirewound ballast resistor for maybe 2-3 seconds. But that
neon tube filled at 10 torr of neon will cost nearly as much as a
Chinese co2 tube to be made. Don't EVEN think of using a flourescent
lamp or argon/mercury filled neon, the fill pressure is too low to
have a positive enough IV curve to match a co2.
I've been working with co2 supplies since 1990, and high voltage
plasma supplies since 1980 or so, I assure you, the only load that psu
will like is a co2 tube with a ballast resistor.
Steve
Neil Baylis
>
> 1. You DONT. running open or shorted will KILL the unit. Running
> without a plasma load of some sort will kill the unit, as will any
> ARCING.
> The Chinese switchers simply do not have the protection systems a
> western power supply does.
What happens when the tube dies in a way that renders it non-
conductive? Will that kill the PSU
as well? Or maybe I should ask, when they die, do they become
instantly non-conductive?
What's the best way to connect the HV wires to the tube? I found some
nice socket terminations
that grip the tube electrodes pretty firmly, and I was planning to
solder those to the PSU leads &
cover with heat shrink or teflon tubing. Is there a better method? I'm
assuming that arcing in these contacts
could also damage the PSU, yes? Or did you just mean arcing to ground?
>
> 2. You could test it on a neon tube about 3 foot long with a 50-80K 5
> watt series wirewound ballast resistor for maybe 2-3 seconds. But that
> neon tube filled at 10 torr of neon will cost nearly as much as a
> Chinese co2 tube to be made. Don't EVEN think of using a flourescent
> lamp or argon/mercury filled neon, the fill pressure is too low to
> have a positive enough IV curve to match a co2.
No, this doesn't sound like it's worth the effort, although it might
look cool.
But you mentioned ballast resistors. The hookup instructions say to
connect the PSU
directly to the tube, with no ballast resistor. This implies to me
that the PSU is a
current regulator. They say it has closed loop output control.
Today I'm going to bring the PSU to work, and have a Chinese colleague
translate
the chinese text on the outside, and then I'll open it up to see if
there's a ballast resistor
inside, but I think there isn't.
>
> I've been working with co2 supplies since 1990, and high voltage
> plasma supplies since 1980 or so, I assure you, the only load that psu
> will like is a co2 tube with a ballast resistor.
>
> Steve
Thanks for the info, very helpful.
Neil
Neil Baylis
somehow my previous reply to your post vanished. Trying again:
>
> First thing is to read and think and sit and watch wait for better replies,
> etc.. and do anything other than try it till you have no further reason NOT
> to do so, and you might find a proper laser tube load before you try this PSU
> some other way.
I do have a small tube I could use, but it's been sitting on the shelf
for years, and
probably doesn't work. Would I do more harm than good by attempting to
test the PSU
with a questionable tube?
> If you have 30 mA being pushed around by 30 kV you have such a high risk of
> immediate painful death that playing Russian Roulette by dodging the
> raindrops from a lawn sprinkler spurting concentrated H2SO4 can be considered
> the soft option! There is no way to overstate this case so I might as well
> wax lyrical about it. No matter what you know about HV, reconsider it deeply
> before you try that thing.
I understand, thanks for the warning.
>
> If you're really set on trying this, then I suggest using a
> series chain of solid carbon composite resistors. While they vary a lot of
> value under any conditions let alone these, they won't burn so fast you can't
> react in time, and likely won't burn at all if you calculate right. Make the
> network on a clean dry slate or ceramic slab. Trying to suspend it in air
> might be electrically safe IF it holds under stress, but physically so
> accident prone I wouldn't dare try it.
Yes, this is more or less what I was planning. I was going to put a
100mA
meter at the 'bottom' of the chain, and also measure the voltage
across the 'bottom'
resistor. I could potentially add or subtract resistors at the HV end
to change
the load on the PS and see how it responds.
>
> When you say 'do 40W' I take it you mean it will do enough to push 40W out of
> a tube, plus whatever gets lost from that which gets pushed in?
Yes, 40 watts out. There's a couple of hundred Watts going into the
water jacket. But I just used that tube as an
example. I'll likely be using a tube that puts out maybe 20 Watts.
I'm building a microcontroller to manage the cooling and safety
interlocks, and I want to be sure it's working
correctly.
Thanks for your help.
Neil
Lostgallifreyan
> Hi Doc,
>
> somehow my previous reply to your post vanished. Trying again:
>
It might sahow up 5 days from now. :) Seen a lot of that on Usenet in the
last month or two.
>> If you're really set on trying this, then I suggest using a
>> series chain of solid carbon composite resistors. While they vary a lot
>> of value under any conditions let alone these, they won't burn so fast
>> you can't react in time, and likely won't burn at all if you calculate
>> right. Make the network on a clean dry slate or ceramic slab. Trying to
>> suspend it in air might be electrically safe IF it holds under stress,
>> but physically so accident prone I wouldn't dare try it.
>
> Yes, this is more or less what I was planning. I was going to put a
> 100mA
> meter at the 'bottom' of the chain, and also measure the voltage
> across the 'bottom'
> resistor. I could potentially add or subtract resistors at the HV end
> to change
> the load on the PS and see how it responds.
>
Since I posted, Steve Roberts posted, and I see you've seen that and replied
too. While a series chain of carbon composite resistors carefully calculated
will protect you by not burning it won't mimic the relation of volts and amps
in an actual CO2 laser tube. If the PSU is as limited in protections as Steve
suggests, then definitely hold out for an actual tube. Given that a cheap PSU
made for one specific purpose would have been designed and tested only in
that special case, the safety margins might be too small to trust. it might
even break in ways that make it dangerous even if you did use resistors that
might otherwise be safe. Internal arcing, perhaps. Could be explosive in a
potted system...
> I'm building a microcontroller to manage the cooling and safety
> interlocks, and I want to be sure it's working
> correctly.
>
Wise. I imagine that can be done ok even without the actual PSU operating
though, unless you're specifically monitoring for presence of HV or laser
radiation, and you can simulate those, probably.
Lostgallifreyan
b3ee19...@f20g2000prn.googlegroups.com:
> But you mentioned ballast resistors. The hookup instructions say to
> connect the PSU
> directly to the tube, with no ballast resistor. This implies to me
> that the PSU is a
> current regulator. They say it has closed loop output control.
>
> Today I'm going to bring the PSU to work, and have a Chinese colleague
> translate
> the chinese text on the outside, and then I'll open it up to see if
> there's a ballast resistor
> inside, but I think there isn't.
>
That might be what Steve was hinting at, because if there is current control
via a sense resistor that isn't also a ballast resistor, a failure of control
could result in a wild error in output. That might relate to your earlier
question about what happens if a tube fails open-circuit, in which case (and
hopefully ALL cases) the PSU dies and must be replaced along with the tube
but it at least dies in a safe way whatever happens. I imagine there are
limits to how badly corners are cut, as people don't do repeat business with
firms whose products burn labs down when they fail, or kill their operators,
and it appears that to make the product cheap they probably do depend on a
lot of repeat business in replacement PSU's and tubes because the market for
them isn't big like it is for computer PSU's... Unless you get one tested in
strict conditions it might be hard to know the answer to that. Experience
from people who have seen a lot of them fail is likely as good as you'll
find. I'll stop now because I have never actually had one. :)
Samuel M. Goldwasser
There almost certainly needs to be a ballast resistor near the tube just
like with a HeNe. Except that for a CO2 laser, the ballast resistance
needs to be much higher to maintain stability. I've seen 200K
So, when the pwoer supply says to connect to the tube, that may simply
be a poor translation. It would take a might sophisticated regulation
scheme to avoid the need for a passive ballast of some kind, if even possible.
Note that even trying a purely resistive load may not prove anything
unless (1) it works or (2) blows up. :) Many HeNe inverter-type power
supplies will not start up into a resistive load, even a very high value
one.
--
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Neil Baylis
> Lostgallifreyan <no-...@nowhere.net> writes:
> > b3ee19b2c...@f20g2000prn.googlegroups.com:
It's strange. I opened up the power supply, and there's no ballast
resistor inside. The HV
output comes directly out of an inductor that looks a bit like a
flyback transformer. That's
a thick red cable that you connect to the anode of the tube. The
return side is a thin black wire
that's just soldered onto the PCB and goes to ground through 22 Ohms.
Interestingly, 22 Ohms x 30 mA is 0.6 Volts, so maybe there's an
overcurrent protection circuit there.
I've contacted folks on other forums that deal with these lasers, to
see if anyone uses a
ballast resistor, but I think the answer is going to be no.
I've also uploaded the instructions that came with mine so folks can
see, if interested.
http://www.3worlds.net/psu.doc
It's a little hard to decipher, but the diagram is unambiguous, and
shows no resistor.
In the doc, when they say electricity, they mean current. There's an
analog input that
controls the power (I think by regulating the current). You can feed
0-5 Volts analog signal in,
or a 0-5V PWM digital signal, so it's just an analog input with an
LPF.
Inside the power supply, there are a couple of large anonymous potted
blocks full of magic that
control the supply. One of them has two trimpots poking out. One
trimpot adjusts the threshold
for the digital input signals. The other seems to adjust the trigger
voltage for starting the tube.
Anyway, I guess this is mostly academic now. I've more or less given
up on the dummy load idea,
because everyone seems to think it's unwise.
Neil Baylis
current regulation, and so no ballast resistor is needed.
Lostgallifreyan
f1686a...@u18g2000pro.googlegroups.com:
> I verified with folks on another forum that these PSUs do closed loop
> current regulation, and so no ballast resistor is needed.
Interesting. I suspected it was doable (though at same time agreeing with Sam
that it was unlikely). Sam said a ballast resistor would be needed for
stability, but I imagine that this isn't always so, but not having one leaves
a very narrow safety margin and strongly demands agility and accuracy in the
control loop, but as that's what they are doing, it means that tiny errors in
the loop can cause wild excursions in output because the sense resistor is
likely a small value with small voltage across (relatively) it. I suppose it
comes down to the cost of the PSU. Is it worth paying for a new one to find
out if the existing one can cope with an unintended load. Whatever it lacks
in protection, if it was designed to survive a tube failure you do have
something to go on, just by finding out how those tubes usually fail.
James Sweet
15KV is the open circuit voltage, and 30mA is the current limit. In
actual operation, the voltage will be significantly lower.
As someone else suggested, you could test it with a neon tube, the
characteristics are similar. Personally though I would just hook it up
to the laser tube.
STEVE ROBERTS
resistor.
I have seen that before.
Steve
Neil Baylis
> Personally though I would just hook it up
> to the laser tube.
Yeah, I think that's what I'll end up doing. It's a hassle, because
it means I have to have cooling up and running as well. But I have
to do that eventually anyway, so what the hell.
Neil
Samuel M. Goldwasser
You could run it for a few seconds without cooling just to see if it lights
and is stable.
James Sweet
Just hook up a hose and get water flowing through, you can use a total
loss system for testing, one hose to the tap and another back to the
sink drain.
James Sweet
> Neil Baylis <neil....@gmail.com> writes:
>
>> On Dec 20, 1:59 pm, James Sweet <jamesrsw...@gmail.com> wrote:
>>
>>
>>> Personally though I would just hook it up
>>> to the laser tube.
>> Yeah, I think that's what I'll end up doing. It's a hassle, because
>> it means I have to have cooling up and running as well. But I have
>> to do that eventually anyway, so what the hell.
>
> You could run it for a few seconds without cooling just to see if it lights
> and is stable.
>
At the very least I would fill up the jacket with water. Best to just
get it flowing properly anyway though.